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  • A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of Triangle ADE.

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of \triangleADE.

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Solution.
Distance \, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
The given points A(6, 1), B(8, 2) and C(9, 4) let D(x, y)

As the diagonal of a parallelogram bisect each other.
Here mid-point of AC = mid-point of BD
\left ( \frac{6+9}{2},\frac{1+4}{2} \right )= \left ( \frac{8+x}{2},\frac{2+y}{2} \right )
\left ( \frac{15}{2},\frac{5}{2} \right )= \left ( \frac{8+x}{2},\frac{2+y}{2} \right )
\frac{15}{2}= \frac{8+x}{2}       \frac{2+y}{2}= \frac{5}{2}

x = 7         y = 3
D (7, 3)
E is the mid-point of CD
Let E(x0, y0)
\left ( x_{0},y_{0} \right )= \left ( \frac{7+9}{2}, \frac{3+4}{2} \right )
\left ( x_{0},y_{0} \right )= \left ( \frac{16}{2},\frac{2}{7} \right )
E= \left ( 8,\frac{7}{2} \right )
Area \, of \, \triangle ADE = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
                                    = \frac{1}{2}\left [ 6\left ( \frac{7}{2}-3 \right ) +8\left ( 3-1 \right )+7\left ( 1-\frac{7}{2} \right )\right ]
                                     = \frac{1}{2}\left [ 6\times \frac{1}{2}+8\times 2+7\times \frac{-5}{2}\right ]
                                      = \frac{1}{2}\left [ 3+16-\frac{35}{2}\right ]
                                      = \frac{1}{2}\left [ \frac{6+32-35}{2}\right ]
Area\, of\, \triangle ADE= \frac{1}{2}\times \frac{3}{2}= \frac{3}{4}\, sq\cdot units 

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infoexpert27

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